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My exam review states that I need to utilize the difference formula for sine to solve the equation on the interval $0 \leq \theta < 2\pi $

$$\sqrt3\sin \theta- \cos\theta = 1$$

I know that: $\sin \frac\pi3 = \frac{\sqrt3}{2}$ and $\cos\frac\pi3 = \frac12 $, so I divide each term by 2 and rewrite the equation:

$$\frac{\sqrt3\sin\theta}{2} - \frac{\cos\theta}{2} = \frac12$$

from here, I apply the difference formula for sine:

$$\sin(\alpha-\beta) = (\sin\alpha \cdot \cos\beta) - (\cos\alpha \cdot \sin\beta)$$

(this is the step that I believe I'm doing incorrectly. I've tried plugging in the corresponding numerical values for sin/cos into the equation instead, but I was still unable to go past this step.)

$$\sin(\alpha-\beta) = \left(\frac {\pi}{3} \cdot \frac{\pi}{3}\right) - \left(\frac{\pi}{6} \cdot \frac{\pi}{6}\right) = \frac12$$

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5 Answers 5

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$$\sqrt3\sin \theta- \cos \theta = 1$$

$$2(\frac{\sqrt3}{2} \sin \theta- \frac{1}{2}\cos \theta)=1$$

$$(\frac{\sqrt3}{2} \sin \theta- \frac{1}{2}\cos \theta)=\frac12$$

$$ (\cos \dfrac{\pi}{6}\sin \theta -\sin \dfrac{\pi}{6} \cos \theta) = \sin \dfrac{\pi}{6} $$

$$\sin\left(\theta-\dfrac\pi6\right)=\sin\dfrac\pi6$$

$$\left(\theta-\dfrac\pi6\right)=\dfrac\pi6,\, \pi-\dfrac\pi6 .... $$

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What you are missing is how to match the addition formula and the given equation.

You have $$\sin(\alpha-\beta)=\sin\alpha\cos\beta-\cos\alpha\sin\beta$$ vs. $$\sin \theta\cdot\frac{\sqrt3}2-\cos\theta\cdot\frac12.$$

Then you identify

$$\alpha\leftrightarrow\theta\text{ and }\beta\leftrightarrow \frac\pi6$$

and get

$$\sin\left(\theta-\frac\pi6\right)=\frac12.$$

You should be able to finish.

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    $\begingroup$ If $\cos\beta = \frac{\sqrt{3}}{2}$ and $\sin\beta = \frac{1}{2}$, then $\beta = \frac{\pi}{6}$. $\endgroup$ Commented Apr 2, 2017 at 11:48
  • $\begingroup$ @N.F.Taussig: yep, I had $50\%$ chances. $\endgroup$
    – user65203
    Commented Apr 2, 2017 at 12:07
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$$\sin\left(\theta-\dfrac\pi6\right)=\sin\dfrac\pi6$$

Now $\sin x=\sin A,x=n\pi+(-1)^nA$ where $n$ is any integer

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$$\sqrt3\sin \theta- \cos \theta = 1$$

$$\frac{\sqrt3}{2} \sin \theta- \frac{1}{2}\cos \theta = \frac{1}{2}$$

Now, $\sin \dfrac{\pi}{6}=\dfrac{ 1}{2}$ and $\cos \dfrac{\pi}{6}=\dfrac{ \sqrt{3}}{2}$

$$\cos \dfrac{\pi}{6} \sin \theta- \sin \dfrac{\pi}{6}\cos \theta = \frac{1}{2}$$

Let $\dfrac{\pi}{6}=A$ $$\sin(\theta -A) = \sin \theta \cos A - \cos \theta \sin A$$

$$\sin \Big(\theta - {\dfrac{\pi}{6}}\Big) =\cos \dfrac{\pi}{6} \sin \theta- \sin \dfrac{\pi}{6}\cos \theta = \frac{1}{2}$$

$$\implies \Big(\theta - {\dfrac{\pi}{6}}\Big)= n\pi+(-1)^n\frac{\pi}{6}$$

$$\implies \theta = \Big(n+\frac{1}{6} \Big)\pi+(-1)^n\frac{\pi}{6}$$

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Approach this systematically as follows. Express your LHS (left-hand side) as $R\sin(\theta-B)$ where you need to find $R$ and $B$. So $$ R\sin(\theta-B)=R (\sin(\theta)\cos(B)-\cos(\theta)\sin(B)) $$ Comparing with $\sin(\theta)$ and $\cos(\theta)$ on your LHS we find $$ R\cos(B)=\sqrt3 $$

$$ R\sin(B)=1 $$ Now square and add these $$ R^2=4 \Rightarrow R=2 $$ and divide these $$ \tan{B}=\frac{1}{\sqrt3} \Rightarrow B=\pi/6 $$ Thus $$ 2\sin(\theta-\pi/6)=1 \Rightarrow \sin(\theta-\pi/6)=1/2 $$ Finally $$ \theta-\pi/6 = n\pi+(-1)^n\pi/6 $$

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  • $\begingroup$ I'd be pleased to hear why this is not useful. $\endgroup$
    – PM.
    Commented Apr 2, 2017 at 10:58
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    $\begingroup$ I did not understand basis of negative voting.. $\endgroup$
    – Narasimham
    Commented Apr 2, 2017 at 12:05
  • $\begingroup$ @Narasimham i appreciate your comment, thanks. $\endgroup$
    – PM.
    Commented Apr 2, 2017 at 12:10
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    $\begingroup$ @scitamehtam Thank you for taking the time to help. It's much appreciated. I upvoted this as soon as I became aware of it. $\endgroup$
    – JCD
    Commented Apr 2, 2017 at 16:19
  • $\begingroup$ @JCD thanks and you're very welcome $\endgroup$
    – PM.
    Commented Apr 2, 2017 at 18:05

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